Permutations and Combinations

More Topic from Mathematics

Question 1

How many different words can be formed by jumbling the letters in the word MISSISSIPPI in which no two S are adjacent?

  • 8 . 6C4 . 7C4

  • 6 . 7 . 8C4

  • 6 . 8 . 7C4

  • 7 . 6C4 . 8C4

Solution

D.

7 . 6C4 . 8C4

Other than S, seven letters M, I, I, I, P, P, I can be arranged in 7!/2! 4!=7 . 5 . 3.
Now four S can be placed in 8 spaces in 8 C4 ways. Desired number of ways = 7 . 5 . 3 . 8C4 = 7 . 6C4 . 8C4.

Question 2

How many ways are there to arrange the letters in the word GARDEN with the vowels in alphabetical order?

  • 120

  • 480

  • 360

  • 240

Solution

C.

360

A total number of ways in which all letters can be arranged in alphabetical order = 6! There are two vowels in the word GARDEN. A total number of ways in which these two vowels can be arranged = 2!
∴ Total number of required ways 
∴ Total number of required waysspace equals space fraction numerator 6 factorial over denominator 2 factorial end fraction space equals space 360

Question 3

If m is the AMN of two distinct real numbers l and n (l,n>1) and G1, G2, and G3 are three geometric means between l and n, then straight G subscript 1 superscript 4 space plus 2 straight G subscript 2 superscript 4 space plus space straight G subscript 3 superscript 4 equals

  • 4l2 mn

  • 4lm2n

  • 4 lmn2

  • 4l2m2n2

Solution

B.

4lm2n

Given, 
m is the AM of and n
l +n = 2m
and G1, G2, G3, n are in GP
Let r be the common ratio of this GP
G1 = lr
G2 =lr2
G3= lr3
n = lr4
rightwards double arrow space straight r space equals space open parentheses straight n over l close parentheses to the power of 1 divided by 4 end exponent
Now comma space straight G subscript 1 superscript 4 space plus space 2 straight G subscript 2 superscript 4 space plus space straight G subscript 3 superscript 4 space equals left parenthesis l straight r right parenthesis to the power of 4 space plus space left parenthesis l straight r squared right parenthesis to the power of 4 space plus space left parenthesis l straight r cubed right parenthesis to the power of 4
space equals space straight l to the power of 4 space straight x space straight r to the power of 4 left parenthesis 1 plus 2 straight r to the power of 4 plus straight r to the power of 8 right parenthesis
equals space straight l to the power of 4 space straight x space straight r to the power of 4 space left parenthesis straight r to the power of 4 space plus 1 right parenthesis squared
equals space straight l to the power of 4 space straight x space straight n over straight l open parentheses fraction numerator n italic plus italic 1 over denominator l end fraction close parentheses
equals space l n italic space x italic 4 m to the power of italic 2 italic space italic equals italic space italic 4 l m to the power of italic 2 n

Question 4

The number of integers greater than 6000 that can be formed, using the digits 3,5,6,7 and 8 without repetition, is

  • 216

  • 192

  • 120

  • 72

Solution

B.

192

Question 5

The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is

  • 5

  • 8C3
  • 38

  • 21

Solution

D.

21

The required number of ways
space equals space to the power of 8 minus 1 end exponent straight C subscript 3 minus 1 end subscript
space equals space to the power of 7 straight C subscript 2 space equals space fraction numerator 7 factorial over denominator 2 factorial 5 factorial end fraction
space equals space to the power of 7 straight C subscript 2 space equals space fraction numerator 7 space factorial over denominator 2 factorial space 5 factorial end fraction
space equals space fraction numerator 7.6 over denominator 2.1 end fraction space equals space 21